Question: $\int^{\pi/3}_{0}\sin(x)\cos^3(x)\,dx\, = $
Answer: Strategy Let's first find the indefinite integral $\int\sin(x)\cos^3(x)\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\sin(x)\cos^3(x)\,dx\, $, we can use U-substitution. If we let $ {u=\cos(x)}$, then ${du=-\sin(x) \, dx}$ and ${ \sin(x)\,dx=-du}$. So we have: $\begin{aligned}\int\sin(x)\cos^3(x)\,dx\,&=\int{\cos}^3{(x)}{\sin(x)\, dx}\,\\\\\\\\ &=\int u^3\,\cdot {-du} \,\\\\\\\\ &=-\int u^3\,\cdot du \,\\\\\\\\ &=-\dfrac{1}{4}\cdot u^{4}+C\\\\\\\\ &=-\dfrac14\cos^4(x)+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{\pi/3}_{0}\sin(x)\cos^3(x)\,dx\,&= -\dfrac14\cos^4(x)\Bigg|^{\pi/3}_0\\\\\\\\ &=-\dfrac{1}{4}\left(\cos^4\left(\dfrac{\pi}{3}\right)-\cos^4(0)\right)\\\\\\\\ &=-\dfrac{1}{4}\left(\left(\dfrac12\right)^4-1^4\right)\\\\\\\\ &=-\dfrac14\cdot-\dfrac{15}{16}\\\\\\\\ &=\dfrac{15}{64}\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{\pi/3}_{0}\sin(x)\cos^3(x)\,dx\, = \dfrac{15}{64}$